Solve -3x+1/2x-1geq1

$$- \frac { 3 x + 1 } { 2 x - 1 } \geq 1$$

Solve for x

$$x\in [0,\frac{1}{2})$$

Solution Steps

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$\frac{3x+1}{2x-1}\leq -1$$

Denominator 2x-1 cannot be zero since division by zero is not defined. There are two cases.

$$2x-1>0 2x-1<0$$

Consider the case when 2x-1 is positive. Move -1 to the right hand side.

$$2x>1$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x>\frac{1}{2}$$

The initial inequality does not change the direction when multiplied by 2x-1 for 2x-1>0.

$$3x+1\leq -\left(2x-1\right)$$

Multiply out the right hand side.

$$3x+1\leq -2x+1$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$3x+2x\leq -1+1$$

Combine like terms.

$$5x\leq 0$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

$$x\leq 0$$

Consider condition x>\frac{1}{2} specified above.

$$x\in \emptyset$$

Now consider the case when 2x-1 is negative. Move -1 to the right hand side.

$$2x<1$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x<\frac{1}{2}$$

The initial inequality changes the direction when multiplied by 2x-1 for 2x-1<0.

$$3x+1\geq -\left(2x-1\right)$$

Multiply out the right hand side.

$$3x+1\geq -2x+1$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$3x+2x\geq -1+1$$

Combine like terms.

$$5x\geq 0$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

$$x\geq 0$$

Consider condition x<\frac{1}{2} specified above.

$$x\in [0,\frac{1}{2})$$

The final solution is the union of the obtained solutions.

$$x\in [0,\frac{1}{2})$$